Explicit Arithmetic Of Modular Curves - Lecture 2
Weil Pairing
There is a Weil pairing on the group of \(N\)-torsion points \(E[N]\) of an elliptic curve \(E\) over a perfect field \(K\) of characteristic \(\mathrm{char}K \nmid N\). (This makes sure \(E[N]\cong \mathbb{Z}/N\mathbb{Z}\times \mathbb{Z}/N\mathbb{Z}\), in bad characteristic, the torsion group can be smaller.)
\[ e_N : E[N] \times E[N] \to \mu_N \]
where \(\mu_N\) is the group of \(N\)-th roots of unity. This is a
bilinear
\[ e_N(aP+bQ, R) = e_N(P, R)^a e_N(Q, R)^b \]
non-degenerate
\[ S\in E[N] \quad e_N(P, S) = 1 \quad \forall P \in E[N] \Rightarrow S = 0 \]
alternating
\[ e_N(P, P) = 1 \quad \forall P \in E[N] \]
together with bilinear, this implies \(e_N(T,S) = e_N(S,T)^{-1}\).
Galois equivariant
\[ e_N(\sigma P, \sigma Q) = \sigma e_N(P, Q) \]
for all \(\sigma \in \mathrm{Gal}(\overline{K}/K)\).
Theorem \(\det\circ \overline{\rho}_{E,N} = \chi_N \in \mathrm{Hom}(G_K, (\mathbb{Z}/N\mathbb{Z})^\times)\).
Recall that we have a morphism telling me what \(G_K\) is doing to my torsion points
\[G_K \xrightarrow{\overline{\rho}_{E,N}} \mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z}) \xrightarrow{\det} (\mathbb{Z}/N\mathbb{Z})^\times\]
when I take the composition, I get a multiplicative character \(\chi_N\). Think it as the one dimensional representation of \(G_K\).
When I take a primitive root \(\zeta_N\), and we apply a Galois action, we will get another primitive root \(\zeta_N^\sigma = \zeta_N^{\chi_N(\sigma)}\).
Definition. A symplectic basis for \(E[N]\) is a \(\mathbb{Z}/N\mathbb{Z}\) basis \(S, T\) satisfying \(e_N(S,T) = \zeta_N\).
Lemma. Let \(S\in E[N]\) have order exactly \(N\), then it can be extended to a symplectic basis \((S, T)\).
Consider the map \(\psi_S = e_N(S,\cdot) : E[N] \to \mu_N\), by non-degeneracy of the Weil pairing the image of \(\psi_S\) have size \(N\), since otherwise it is annihilated by a proper factor of \(N\), which contradicts the order of \(S\).
The kernel of \(\psi_S\) is just \(\langle S \rangle\).
Proof of Theorem. Let \(S,T\) be a symplectic basis for \(E[N]\), then we can take an element \(\sigma\in G_K\) and write \(\overline{\rho}_{E,N}(\sigma)\) in the basis \((S,T)\).
\[\overline{\rho}_{E,N}(\sigma) = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
i.e. this means \(S^\sigma = aS + cT\) and \(T^\sigma = bS + dT\).
Then we can compute
\[\begin{aligned} \zeta_N^{\chi_N(\sigma)} &= \zeta_N^\sigma \\ &= e_N(S^\sigma, T^\sigma) \\ &= e_N(aS + cT, bS + dT) \\ &= e_N(S, T)^{ad-bc} \\ &= \zeta_N^{ad-bc} \\ &= \zeta_N^{\det(\overline{\rho}_{E,N}(\sigma))} \end{aligned}\]
This shows that \(\det(\overline{\rho}_{E,N}) = \chi_N\) as functions taking value in \(\mathbb{Z}/N\mathbb{Z}\).
Definition Suppose \(K\) is a number field, we say that \(\overline{\rho}:G_K\to \mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})\) is odd if \(\det(\overline{\rho}(\tau)) = -1\) for all complex conjugations \(\tau\in G_K\).
If I have an embedding \(K\hookrightarrow \mathbb{R}\) I can always extend it to an embedding \(\overline{K}\hookrightarrow \mathbb{C}\)
we have \(\chi_N(\tau) = -1\) for all complex conjugations \(\tau\in G_K\). As a corollary of the theorem, \(\overline{\rho}_{E,N}\) is odd.
Theorem The following are equivalent
\(E\) have a \(K\)-rational point of order \(N\).
\(\overline{\rho}_{E,N}\sim \begin{pmatrix} 1 & * \\ 0 & \chi_N \end{pmatrix}\), i.e. I can choose a basis such that the representation looks this, a trivial sub-representation and the obvious character as quotient.
\(\overline{\rho}_{E,N}(G_K)\) is conjugate inside \(\mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})\) to a subgroup of
\[B_1(N) = \left\{ \begin{pmatrix} 1 & b \\ 0 & d \end{pmatrix} : d\in (\mathbb{Z}/N\mathbb{Z})^*, b\in \mathbb{Z}/N\mathbb{Z} \right\} \]
Proof
1 -> 2.
Let \(P\) be a \(K\) rational point of order \(N\), extend \(P\) to \(P,Q\) as a symp basis for \(E[N]\), let \(\sigma\in G_K\) and now, by rationality of \(P\)
\[ P^\sigma = P + 0Q, \quad Q^\sigma = bP + dQ \]
we have a matrix form
\[ \begin{pmatrix} 1 & b \\ 0 & d \end{pmatrix} \]
where \(d = \det(\overline{\rho}_{E,N}(\sigma)) = \chi_N(\sigma)\).
2 -> 3 is obvious.
3 -> 1. The existence of such a basis implies rationality of the first basis vector.
This tells us we can detect rational torsion points from the representation!
Isogenies
For convenience we assume \(\mathrm{char}K= 0\), so we don’t need to distinguish separable and inseparable isogenies.
Let \(E,E'/K\) be two elliptic curves, a \(K\)-isogeny \(\phi:E\to E'\) is a non-constant morphism defined over \(K\) that is also a group homomorphism, i.e. satisfying \(\phi(0)=0\). (A theorem tells us that preserving \(0\) is equivalent to being a group homomorphism for elliptic curves.)
The degree of \(\phi\) is the degree of the field extension \([K(E):\phi^*K(E')]\), which equals the size of kernel of \(\phi\) is a finite subgroup of \(E\).
Since \(\phi\) is defined over \(K\), the preimage of any \(K\)-point consisting of \(\deg(\phi)\) points that is Galois closed, e.g. \(\ker(\phi)\) is a Galois module.
\(\phi\) is called cyclic if \(\ker(\phi)\) is cyclic.
Theorem. TFAE
\(E\) has a cyclic \(K\)-isogeny of degree \(N\)
\[\overline{\rho}_{E,N}\sim \begin{pmatrix} \phi & * \\ 0 & \psi \end{pmatrix} \]
where \(\phi, \psi:G_K\to (\mathbb{Z}/N\mathbb{Z})^\times\) are characters.
\(\overline{\rho}_{E,N}\) is conjugate inside \(\mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})\) to a subgroup of the form
\[B_0(N):= \left\{ \begin{pmatrix} a & b \\ 0 & d \end{pmatrix} : a,d\in (\mathbb{Z}/N\mathbb{Z})^*, b\in \mathbb{Z}/N\mathbb{Z} \right\} \]
Proof.
1 -> 2. \(\ker(\phi)\subset E[N]\) is cyclic of order \(N\) by definition,
we can write \(\ker(\phi) = \langle P\rangle\) where \(P\) has order \(N\). Extend \(P\) to \(P,Q\) a basis of \(E[N]\) we can write
\[P^\sigma = a_\sigma P + 0Q, \quad Q^\sigma = b_\sigma P + d_\sigma Q\]
\[\overline{\rho}_{E,N}(\sigma) = \begin{pmatrix} a_\sigma & b_\sigma \\ 0 & d_\sigma \end{pmatrix} \]
define \(\eta,\psi:G_K\to (\mathbb{Z}/N\mathbb{Z})^\times\) by \(\eta(\sigma) = a_\sigma\), then \(\eta\) is a homomorphism. Similarly define \(\psi\).
Quadratic Twisting
Lemma. Let \(d\in K^*\), suppose \(\mathrm{char}K\neq 2\). Let \(E'\) be the quadratic twist of \(E\) by \(d\). Let \(\psi:G_K\to \{\pm 1\}\) be defined by \(\psi(\sigma) = \frac{\sqrt{d}^\sigma}{\sqrt{d}}\). Then
\[\overline{\rho}_{E',N}\sim \psi\cdot \overline{\rho}_{E,N}.\]
Proof. Let \(E: y^2 = x^3+ax^2+bx+c\), then \(E': y^2 = x^3+adx^2+bd^2x+cd^3\).
We can define \(\phi: E(K)\to E'(\overline{K})\) be the isomorphism
\[\phi(x,y) = \left(\frac{x}{d}, \frac{y}{d \sqrt{d}}\right)\]
inducing an isomorphism of \(N\) torsions \(\phi: E[N]\to E'[N]\).
Let \(P,Q\) be a basis for \(E[N]\) so \(\phi(P),\phi(Q)\) is a basis for \(E'[N]\).
\[\begin{align*} \phi(x,y)^\sigma &= \left(\frac{x}{d}, \frac{y}{d \sqrt{d}}\right)^\sigma \\ &= \left(\frac{x^\sigma}{d}, \frac{y^\sigma}{d \sqrt{d}^\sigma}\right) \\ &= \left(\frac{x^\sigma}{d}, \frac{y^\sigma}{d \psi(\sigma)\sqrt{d}}\right) \\ &\quad \text{because } -(x,y) = (x, -y) \\ &= \psi(\sigma) \left(\frac{x^\sigma}{d}, \frac{y^\sigma}{d \sqrt{d}}\right) \\ &= \psi(\sigma) \phi(x^\sigma, y^\sigma) \end{align*}\]
This tells us
\[\psi(\sigma)\phi(P^\sigma) = \phi(P)^\sigma\]
\[\psi(\sigma)\phi(Q^\sigma) = \phi(Q)^\sigma\]
Let \(P^\sigma = aP + cQ, Q^\sigma = bP + dQ\), then
\[\phi(P)^\sigma = \psi(\sigma) \phi(P) = \psi(\sigma) (a\phi(P) + c\phi(Q))\]
\[\phi(Q)^\sigma = \psi(\sigma) \phi(Q) = \psi(\sigma) (b\phi(P) + d\phi(Q))\]
\[\overline{\rho}_{E,N}(\sigma) = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \quad \overline{\rho}_{E',N}(\sigma) = \psi(\sigma) \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
Theorem Let \(H\) be a subgroup of \(\mathrm{GL}_2(\mathbb{Z}/N)\). Suppose \(\overline{\rho}_{E,N}(G_K)\subset H\) and \(-I\in H\). Let \(E'\) be a quadratic twist of \(E\) then \(\overline{\rho}_{E',N}(G_K)\subset H\).
Since as long as \(-1\in H\), the representation is preserved under quadratic twisting.
Corollary. If \(E\) has a cyclic \(K\)-isogeny of degree \(N\) and \(E'\) is a quadratic twist of \(E\), then \(E'\) has a cyclic \(K\)-isogeny of degree \(N\).
Since we can pick \(H = B_0(N)\), and \(-I\in B_0(N)\).
(This proof does not work with \(K\)-rational points, because \(B_1(N)\) is not stable under quadratic twisting.)